Question

Mechanism of a hypothetical reaction \(X_2 + Y_2→ 2XY\) is given below :
(i) \(X_2→ X + X\) (fast)
(ii) \(X + Y_2⇋ XY + Y\) (slow)
(iii) \(X + Y → XY\) (fast)
The overall order of the reaction will be

• A. 1
• B. 2
• C. 0
• D. 1.5

Answer 1

The Correct Option is D

To determine the overall order of the reaction, we must consider the given reaction mechanism and utilize the concept of rate-determining step in reaction kinetics.

The given reaction mechanism is:

1. \(X_2 \rightarrow X + X\) (fast)
2. \(X + Y_2 \rightleftharpoons XY + Y\) (slow)
3. \(X + Y \rightarrow XY\) (fast)

Here, the slow step of the reaction, step (ii), is the rate-determining step. The overall rate of reaction is largely dependent on this slow step.

The rate law for the rate-determining step is:

\[\text{Rate} = k_2 [X][Y_2]\]

Where \([X]\) is the concentration of intermediate X and \([Y_2]\) is the concentration of Y_2.

Since X is an intermediate and not a reactant mentioned in the overall reaction, we express the concentration of X in terms of given reactants using the fast pre-equilibrium assumption (step i):

From step (i):

\[K_1 = \frac{[X]^2}{[X_2]}\]

Here K_1 is the equilibrium constant for step (i). We rearrange to express [X]:

\[[X] = \sqrt{K_1 [X_2]}\]

Substitute \([X]\) in the rate law of the rate-determining step:

\[\text{Rate} = k_2 \sqrt{K_1 [X_2]} [Y_2]\]

This can be written as:

\[\text{Rate} = k \cdot [X_2]^{1/2}[Y_2]\]

Here, \(k\) is the composite rate constant, incorporating \(k_2\) and \(K_1^{1/2}\).

The exponents of the concentration terms \([X_2]^{1/2}\) and \([Y_2]^1\) add up to \(0.5 + 1 = 1.5\).

Therefore, the overall order of the reaction is 1.5.